1.

How will you prepare 0.25 m CaVI_(2) solution?

Answer»


Solution :`"MOLALITY of solution"=("No. of moles of"CaCI_(2))/("MASS of SOLVENT in kg ")`
`"No. of moles of "CaCI_(2)="Molality"XX"Mass of solvent in kg"`
`=(0.25" mol lg"^(-1))xx(1 kg )= 0.25 mol`
`"Molar mass of "CaCI_(2)=40+2xx35.5=111 " g mol"^(-1)`
`"Mass of "CaCI_(2)=(0.25 mol)xx(111 g mol^(-1))=27.75 g.`


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